Cho \(S_n=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}\) Tìm các số nguyên dương n sao cho \(\left[S_n\right]=2\)
K
Khách
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18 tháng 2 2018
Câu trả lời ở đây: https://dethihsg.com/de-thi-hoc-sinh-gioi-toan-9-phong-gddt-cam-thuy-2011-2012/amp/
AH
Akai Haruma
Giáo viên
4 tháng 8 2018
Lời giải:
\(S_n=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}+...+\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+..+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Để \(S_n\in\mathbb{Z}\Rightarrow \sqrt{n+1}-1\in\mathbb{Z}\Rightarrow \sqrt{n+1}\in\mathbb{Z}\)
Đặt \(\sqrt{n+1}=t\in\mathbb{N}>1\) do \(n>0\)
\(\Rightarrow n+1=t^2\Rightarrow t^2\leq 101\) do \(n\leq 100\)
\(\Rightarrow 0< t\leq \sqrt{101}\)
Mà \(t\in\mathbb{N}^*\Rightarrow t\in\left\{1;2;3;4;5;6;7;8;9;10\right\}\)
\(\Rightarrow n=t^2-1\in\left\{3; 8; 15; 24;35;48;63;80;99\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 4 2021
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}+1\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^3+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\) \(\Rightarrow a+b=7\)
TT
1
W
16 tháng 1 2018
\(S_n=\sqrt{\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{n}{4^n}}\)
\(S_{16}=\sqrt{\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{16}{4^{16}}}\)
Đặt: \(S_{16}=\sqrt{T}\Leftrightarrow T=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{16}{4^{16}}\)
\(4T=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{16}{4^{15}}\)
\(4T-T=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{16}{4^{15}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{16}{4^{16}}\right)\)
\(3T=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{15}}-\dfrac{16}{4^{16}}\)
Đặt: \(G=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{15}}\)
\(4G=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{14}}\)
\(4G-G=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{14}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{15}}\right)\)
\(3G=4-\dfrac{1}{4^{15}}\)
\(G=\dfrac{4}{3}=\dfrac{1}{4^{15}.3}\)
\(T=\dfrac{4}{3}-\dfrac{1}{4^{15}.3}-\dfrac{16}{4^{16}}\)
\(S_{16}=\sqrt{T}=\sqrt{\dfrac{4}{3}-\dfrac{1}{4^{15}.3}-\dfrac{16}{4^{16}}}\)
16 tháng 1 2018
bn ơi cái này mk bt lm r` sử dụng Xích - ma nha !
kq\(\simeq1,3472\)